Solve stoichiometry problem

Question

1. In dilute nitric acid, HNO3, copper metal dissolves according to the following equation:

Cu(s) + HNO3(aq) → Cu(NO3)2(aq) + NO(g) + H2O(aq)

           How many grams of HNO3 are needed to dissolve 11.45g of Cu?

Answer 1

First we need to balance the equation. As it is a redox reaction, let's write the separate oxidation and reduction reactions.

Cu =====> Cu²⁺ + 2e^-

NO3^1- + 3e^- =======> NO + 2 O^2-

Multiply the first reaction by 3 and the second by 2 to balance the electrons.

3Cu =======> 3Cu + 6e^-

2NO3^1- + 6e^- =========> 2NO + 4O^2-

So, 2 nitrates are required to react with 3 Cu to create 2NO.

A trial balancing gives 3Cu + 2HNO3 =====> 2NO + 3Cu(NO3)2 + H2O

Now we have to balance the extra NO3^- and H2O

There are 6 NO3^- required to balance those in 3Cu(NO3)2, so we have to add 6 more HNO3 to the left side. 3Cu +8HNO3 =====> 2NO + 3Cu(NO3)2 + H2O

Finally we have to balance the H2O. As there are 8H⁺, we need to have 4 H2O giving

3Cu +8HNO3 =====> 2NO + 3Cu(NO3)2 + 4H2O It is finally balanced.

Now form the conversion factor for 3 Cu to 8HNO3. 8 x molecular mass of HNO3 / 3 x atomic mass Cu

= 8 x 63.012 g/mole / 3 x 63.546 g/mole Now multiply by 11.45 g Cu to get 30.3 g HNO3